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# The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint :Sx – 1 = S49 – Sx]

Given,
Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

So,
First term, a = 1
Common difference, d=1

Let us say the number of $$x^{th}$$ houses can be represented as;

Sum of nth term of AP = $${{n} \over {2}}[2a + (n - 1)d]$$
Sum of number of houses beyond x house = Sx-1

= $${{x - 1} \over {2}}[2(1) + (x -1-1)1]$$
= $${{x - 1} \over {2}} [2+x-2]$$
= $${{x(x - 1)} \over {2}}$$………………………………(i)

By the given condition, we can write,

$$S_{49} – Sx = ({{49} \over {2}}[2.1+(49-1)1])–({{x} \over {2}}[2.1+(x-1)1])$$
= $$25(49) – {{x( x + 1)} \over {2}}$$………………………………….(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other;

Therefore,

$${{x(x - 1)} \over {2}} = 25(49) – {{x ( x - 1)} \over {2}}$$
$$x = ±35$$

As we know, the number of house cannot be an a negative number. Hence, the value of x is 35